hdu 1021 斐波那契数列取模（循环节）

http://acm.hdu.edu.cn/showproblem.php?pid=1021

Problem Description

There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).

 

Input

Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).

 

Output

Print the word "yes" if 3 divide evenly into F(n).

Print the word "no" if not.

 

Sample Input

012345

 

Sample Output

nonoyesnonono

找出循环节（8）

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;int a[50];int main(){    int n;    a[0]=7;    a[1]=11;    for(int i=2;i<=8;i++)        a[i]=a[i-1]+a[i-2];    while(~scanf("%d",&n))    {        n=n%8;        n=a[n]%3;        if(n==0)           printf("yes\n");        else           printf("no\n");    }    return 0;}

Problem Description

There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).

 

Input

Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).

 

Output

Print the word "yes" if 3 divide evenly into F(n).

Print the word "no" if not.

 

Sample Input

012345

 

Sample Output

nonoyesnonono